Lemma 56.3.1. Let $A$ and $B$ be rings. Let $F : \text{Mod}_ A \to \text{Mod}_ B$ be a functor. The following are equivalent

1. $F$ is isomorphic to the functor $M \mapsto M \otimes _ A K$ for some $A \otimes _\mathbf {Z} B$-module $K$,

2. $F$ is right exact and commutes with all direct sums,

3. $F$ commutes with all colimits,

4. $F$ has a right adjoint $G$.

Proof. If (1), then (4) as a right adjoint for $M \mapsto M \otimes _ A K$ is $N \mapsto \mathop{\mathrm{Hom}}\nolimits _ B(K, N)$, see Differential Graded Algebra, Lemma 22.30.3. If (4), then (3) by Categories, Lemma 4.24.5. The implication (3) $\Rightarrow$ (2) is immediate from the definitions.

Assume (2). We will prove (1). By the discussion in Homology, Section 12.7 the functor $F$ is additive. Hence $F$ induces a ring map $A \to \text{End}_ B(F(M))$, $a \mapsto F(a \cdot \text{id}_ M)$ for every $A$-module $M$. We conclude that $F(M)$ is an $A \otimes _\mathbf {Z} B$-module functorially in $M$. Set $K = F(A)$. Define

$M \otimes _ A K = M \otimes _ A F(A) \longrightarrow F(M), \quad m \otimes k \longmapsto F(\varphi _ m)(k)$

Here $\varphi _ m : A \to M$ sends $a \to am$. The rule $(m, k) \mapsto F(\varphi _ m)(k)$ is $A$-bilinear (and $B$-linear on the right) as required to obtain the displayed $A \otimes _\mathbf {Z} B$-linear map. This construction is functorial in $M$, hence defines a transformation of functors $- \otimes _ A K \to F(-)$ which is an isomorphism when evaluated on $A$. For every $A$-module $M$ we can choose an exact sequence

$\bigoplus \nolimits _{j \in J} A \to \bigoplus \nolimits _{i \in I} A \to M \to 0$

Using the maps constructed above we find a commutative diagram

$\xymatrix{ (\bigoplus \nolimits _{j \in J} A) \otimes _ A K \ar[r] \ar[d] & (\bigoplus \nolimits _{i \in I} A) \otimes _ A K \ar[r] \ar[d] & M \otimes _ A K \ar[r] \ar[d] & 0 \\ F(\bigoplus \nolimits _{j \in J} A) \ar[r] & F(\bigoplus \nolimits _{i \in I} A) \ar[r] & F(M) \ar[r] & 0 }$

The lower row is exact as $F$ is right exact. The upper row is exact as tensor product with $K$ is right exact. Since $F$ commutes with direct sums the left two vertical arrows are bijections. Hence we conclude. $\square$

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